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Question 125-1 : Planned and actual data as shown in the flight log excerpt..actual ground speed gs on the leg beta to gamma will be 110 kt..if all other flight parameters remain unchanged, what fuel remaining should be expected at waypoint gamma.. err a 033 502 ? [ Preparation civilian ]
2625 kg.
.between alpha to beta, actual flight time was 12 minutes and fuel consumption was 3000 2900 = 100 kg...100 kg / 12 min = 8.33 kg/min...from beta to gamma, distance is 60 nm.60 nm at 110 kt = 60/110 = 0.5454 hour of flight. 8.33 x 60 x 0.5454 = 273 kg...2900 273 = 2627 kg.
Question 125-2 : An aircraft is in cruising flight at fl 095, ias 155 kt. the pilot intends to descend at 500 ft/min to arrive overhead the man vor at 2 000 ft qnh 1 030hpa. the tas remains constant in the descent, wind is negligeable, temperature standard..at which distance from man should the pilot commence the ?
48 nm.
.tas remains constant in the descent and wind is negligeable tas = ground speed...for every 1000 ft of pressure altitude, add 2% to the ias to calculate tas.2% of 155 kt = 155 x 0.02 = 3.1 kt..3.1 x 9.5 = 29.45 kt.tas at fl95 = 155 kt + 29 kt = 184 kt...fl 95 is for 1013 hpa, if we change our subscale settig, we increase indicated altitue by 17 hpa x 27 ft = 459 ft...9500 + 459 2000 = 7959 ft to loose..7959 / 500 = 16 minutes.. 184 /60 x 16 = 49 nm.
Question 125-3 : Given.planned and actual data as shown in the flight log excerpt. arriving overhead gamma you are cleared for direct routing to mike..the flight time for direct flight gamma to mike will be 1h06 min, assuming other flight data remains constant, what fuel will be expected on arrival overhead mike.. ?
1706 kg.
Flight time from alpha to gamma, = 20 minutes + 35 minutes = 55 minutes.fuel used from alpha to gamma = 3400 2630 = 770 kg..fuel flow = 770/55 x 60 = 840 kg/h..gamma to mike 1 h 06 min... 840/60 x 66 = 924 kg...on arrival overhead mike, fuel on board will be.2630 kg 924 kg = 1706 kg.
Question 125-4 : At reference or see flight planning manual mrjt 1 figure 4.7.2. using the above table, in isa conditions and at a speed of m.70/280kias, in an elapsed time of 90 minutes an aircraft with mass at point of diversion 48000 kg could divert a distance of .. err a 033 514 ?
584 nm
Question 125-5 : Flight planning manual mrjt 1 figure 4.7.2.an aircraft on an extended range operation is required never to be more than 120 minutes from an alternate, based on 1 engine inoperative lrc conditions in isa. using the above table and a given mass of 40000 kg at the most critical point, the maximum air ?
735 nm.
. /com en/com033 518.jpg..
Question 125-6 : At reference or see flight planning manual mrjt 1 figure 4.5.3.1.given long range cruise temp. 63° c at fl 330.initial gross mass enroute 54 100 kg. leg flight time 29 min.find fuel consumption for this leg.. err a 033 529 ?
1 100 kg
54100 kg corresponds to 3929 nam..isa 12°c..433 1 kt per degree c below isa = 421 kt..29 minutes at speed 421 kt = 203 nm, 3929 203 = 3726 corresponds to 53000 kg.54100 53000 = 1100 kg...correction isa 12°c = 0.7% ==> 0.7 x 1100 kg = 8 kg....exact answer is 1100 8 = 1092 kg.
Question 125-7 : After flying for 16 min at 100 kt tas with a 20 kt tail wind component, you have to return to the airfield of departure..with a rate turn of 3°/s, you will arrive after ?
25 min.
.180°/3 = 60 secondes for turn back...gs out 100 + 20 = 120 kt..gs home 100 20 = 80 kt...16 x 120/80 = 24 minutes...24 minutes + 1 minute = 25 minutes.
Question 125-8 : At reference or see flight planning manual mrjt 1 figure 4.4.given.twin jet aeroplane, estimated mass on arrival at the alternate 50000 kg.estimated mass on arrival at the destination 52525 kg.alternate elevation msl.destination elevation 1500 ft.find final reserve fuel and corresponding time.. err ?
1180 kg, 30 minutes.
.final reserve fuel is simple calculations based on 30 minutes jet/turbo prop aeroplane hold at endurance speed, calculated with the estimated mass on arrival at the alternate aerodrome or the destination aerodrome, when no destination alternate aerodrome is required eu ops subpart d appendix 1 to 1.255... /com en/com033 560.jpg..2360 / 2 = 1180 kg.
Question 125-9 : In the cruise at fl 155 at 260 kt tas, the pilot plans for a 500 ft/min descent in order to fly overhead man vor at 2 000 feet qnh 1030..tas will remain constant during descent, wind is negligible, temperature is standard..the pilot must start the descent at a distance from man of ?
120 nm.
As we are descending for reaching an altitude, we have to change our subscale from 1013 to 1030 before descent.1013 hpa to 1030 hpa = 17 hpa..we start our descent at an altitude of 15500 + 17 hpa x 27 ft = 15959 ft...15959 ft 2000 ft = 13959 ft to loose..13959 ft / 500 ft/min = 28 minutes...28 min at 260 kt = 28 x 260/60 = 121 nm.
Question 125-10 : For this question use annex 033 11704a.true air speed 170 kt.wind in the area 270°/40 kt.according to the attached the navigation log, an aircraft performs a turn overhead bulen to re route to ard via tgj. the given wind conditions remaining constant..the fuel consumption during the turn is 20 ?
1 545 litres.
Proceed as if there is only one leg, ard to bulen 243 nm..track 123°..fuel consumption from ard to bulen = 869 432 = 437 litres.....on nav computer, we find a 202 kt gso groundspeed out and 135 kt gsh groundspeed home...time on the leg = 243 / 202 = 1.2 h..consumption = 437 / 1.2 = 364 l/h...calculate fuel for the leg bulen ard..time on this leg = 243 / 135 = 1.8 h..consumption = 364 x 1.8 = 655 l....total consumption = 869 + 20 + 655 = 1544 l.
Question 125-11 : At reference 033 345..a twin jet aeroplane, gross mass 200000 kg, begin his cruise leg at the optimum level of mach 0.84 isa, cg=37%, total anti ice devices on. head wind component 30 kt..after 500 nm, to avoid a severe icing area, an immediate descent must be initiated. the airplane mass at the ?
192 500 kg.
On the reference, for 200000 kg we have 6778 nm and 844 min..tas = 484 kt, headwind = 30 kt.gs = 454 kt.time = 500/454 = 1.10h = 66 min.66 min + 7% = 71 min.844 min 71 min = 773 min.on the reference, 773 minutes is corresponding to a value of 192 500 kg... ninorr.where did you take 7% from...on the reference.total anti ice on.fuel= +7%.
Question 125-12 : Planned and actual data as shown in the flight log excerpt 033 145...actual ground speed gs on the leg beta to gamma is 110 kt..if all other flight parameters remain unchanged, what fuel remaining should be expected over gamma.. err a 033 578 ?
2000 kg.
Cqb15 january 2012....between alpha to beta, actual consumption was 2470 2330 = 140 kg.140 kg for 20 minutes column 'ate' , so 140x3 = 420 kg/heure...distance between beta and gamma is 85 nm column 'leg dist' at a ground speed of 110 kt...it will take.85 /110 = 0.773 h...0.773 x 420 = 325 kg of fuel burned...the fuel remaining over gamma should be.2330 325 = 2005 kg..closest answer 2000 kg.
Question 125-13 : The trip fuel for a jet aeroplane to fly from the departure aerodrome to the destination aerodrome is 8350 kg..fuel consumption in holding mode is 2800 kg/h..the quantity of fuel which is needed to carry out one go around and land on the alternate airfield is 4380 kg..the destination aerodrome has a ?
14548 kg.
.minimum quantity of fuel at take off = trip fuel + alternate + contingency + 30 min final reserve jet aircraft...trip fuel = 8350 kg...alternate = 4380 kg...contingency = the greater of 5% of trip or 5 min holding at 1500 ft.5% of trip = 5% x 8350 = 418 kg.5 min holding at 1500 ft = 2800 x 5/60 = 233 kg...30 min final reserve = 2800 /2 = 1400 kg...minimum quantity of fuel at take off = 8350 + 418 + 4380 + 1400 = 14548 kg.
Question 125-14 : Use reference 033 046..a turbojet aeroplane flies using the following data.optimum flight level, mach 0.80.mass 190000 kg.temperature isa.tailwind component 100 kt...fuel mileage and ho y fuel consumption are.... err a 033 580 ?
105 nm/1000 kg.5330 kg/h.
From reference, for 190000 kg, we get 6515 nam and tas 459 kt...6515 459 = 6056 nam > 184 600 kg..190 000 184 600 = 5400 kg/h..distance accomplished in one hour with the wind is 559 nm 459 + 100...thus, fuel consumption for 1000kg is 559/5,4 = 103,5 nm/1000kg.
Question 125-15 : See reference 033 068..given.distance between c to d 540 nm.cruise speed 300 kt ias, at fl210.temperature deviation from isa +20°c.headwind component 50 kt.gross mass at c 60 000 kg..the fuel required from c to d is.. err a 033 581 ?
4240 kg.
Cqb15 january 2012....nam = ground distance x tas/gs.nam = 540 x 406/356 = 616 nm..from reference 033 068, at line 60000, the cruise distance nautical air miles is 3898 nautique air miles. substract 616, you find 3282 nam...back to reference 033 068, what is the mass for 3282 nm we get 55800 kg.60000 55800=4200kg..last step, we have to increase fuel required by 0.5 percent per 10 degrees above isa, we are in isa+20°c, so 1 percent more.4200x1%= 4242 kg....see section 5.4.2 method page 24 on caa cap697 flight planning manual, for that kind of questions...notice you must use the tas given for cruise 406 kt as state in the pdf document.
Question 125-16 : Use route manual chart e hi 2..what is the meaning of the chart information for the beacon s at position 55°59'n 014°06'e.. err a 033 582 ?
Ndb only, ident oe.
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Question 125-17 : Given.planned and actual data as shown in the flight log excerpt. provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged. what fuel remaining should be expected at waypoint delta .. err a 033 587 ?
4940 kg.
Buhoraptor.hello,my answer is 4990 kg..please, can you tell me why there are difference thank you and congratulations for your website....thank you..beta to gamma 10 minutes..fuel used 5440 5340 = 100 kg.fuel flow from beta to gamma 100 / 10 = 10 kg/min.planned time from gamma to delta 1 53 to 2 33 = 40 minutes..planned fuel from gamma to delta 40 min x 10 kg = 400 kg..on arrival overhead delta, fuel on board will be.5340 400 = 4940 kg.
Question 125-18 : Given.planning data as shown in the flight log excerpt fuel planning section..after a balked landing at the destination airport, you have to divert to the alternate airport with the gear extended. the re calculated flight time to the alternate due to the reduced speed is 1h 20min and the fuel flow ?
5669 kg.
.1h40 = 1.33 h.720 x 1,33 = 960 kg..estimated mass at destination est ldg mass at dest. 6629 kg...6629 960 = 5669 kg.
Question 125-19 : Given.planned and actual data as shown in the flight log excerpt..provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged..what fuel remaining should be expected at waypoint delta.. err a 033 589 ?
4550 kg.
.beta to gamma 10 minutes..fuel used 5270 5150 = 120 kg.fuel flow from beta to gamma 120 / 10 = 12 kg/min.fuel used from gamma to delta 50 min x 12 kg = 600 kg..on arrival overhead delta, fuel on board will be.5150 600 = 4550 kg.
Question 125-20 : Given.planned and actual data as shown in the flight log excerpt.provided that flight conditions on the leg gamma to delta remain unchanged and fuel consumption remains unchanged, what fuel remaining should be expected at waypoint delta .. err a 033 590 ?
4690 kg.
.beta to gamma 65 minutes...fuel used 5490 4970 = 520 kg..fuel flow from beta to gamma 520 / 65 = 8 kg/min..fuel used from gamma to delta 35 min x 8 kg = 280 kg....on arrival overhead delta, fuel on board will be..4970 280 = 4690 kg.
Question 125-21 : Vtoss is the take off safety speed for ?
Category a helicopters.
Question 125-22 : Using attached graphic. according to the flight manual diagram, the never exceed speed vne at pressure altitude 10 000 ft with an outside air temperature oat of +10°c and an inflight mass of 2050 kg is.. 687 ?
115 kt.
Vne at 10000 ft, oat +10°c and 2300 kg is 115 kt ias...the graphic states at any gross mass above 2300 kg, decrease vne by 10 kias...since the inflight mass is below 2300 kg, we do not change the value.
Question 125-23 : According to the flight manual diagram, the never exceed speed vne at pressure altitude 10 000 ft with an outside air temperature oat of +10° c and an inflight mass of 2400 kg is.. 662 ?
105 kt.
You must removed 10 kias at any gross mass above 2300 kg.
Question 125-24 : Using attached graphic. which letter indicates the speed for maximum endurance.. 688 ?
B
Power required curve power versus airspeed chart.. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/maximum rate of climb speed/ speed for maximum endurance.
Question 125-25 : Using attached graphic. which letter indicates the speed for the best rate of climb.. 688 ?
B
Power required curve power versus airspeed chart.. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/ maximum rate of climb speed / speed for maximum endurance.
Question 125-26 : Using attached graphic. which letter indicates the speed for best range.. 688 ?
C
The best range will be reach at the minimum total drag speed...power required curve power versus airspeed chart.. 690..a power required to hover outside ground effect..b power required to hover inside ground effect..c translational lift area..d adjustment of power required to counteract sinking..e minimum power for level flight/maximum rate of climb speed/ speed for maximum endurance.
Question 125-27 : A head wind will... ?
Increase the climb flight path angle.
Question 125-28 : A helicopter which has no guaranteed 'stay up' ability in the event of an engine failure is certified in ?
Category b.
No guaranteed stay up ability means that the helicopter cannot maintain level flight in case of an engine failure...category b helicopters single engine or a multi engine helicopter that does not meet category a standards have no guaranteed ability to continue safe flight in the event of an engine failure, and a forced landing is assumed.
Question 125-29 : A helicopter will obtain a maximum flight distance at the speed ?
For maximum range.
Question 125-30 : A hostile sea area is defined as being ?
South of 45°s.
Question 125-31 : A platform is a heliport situated ?
At least 3 m above the surrounding surface
Air operations.def.'elevated final approach and take off area elevated fato ' means a fato that is at least 3 m above the.surrounding surface.
Question 125-32 : A safe forced landing is a landing ?
Unavoidable landing or ditching with a reasonable expectancy of no injuries to persons in the aircraft or on the surface.
Question 125-33 : Aeo means ?
All engines operating.
Question 125-34 : An elevated heliport or helideck is defined as one which is above the surrounding level by ?
3 m.
Air operations.def.'elevated final approach and take off area elevated fato ' means a fato that is at least 3 m above the.surrounding surface.
Question 125-35 : An increase in ambient temperature ?
Generally reduces performance in performance class 1 and especially the take off weight.
Question 125-36 : An increase in density altitude ?
Generally reduces performance in performance class 1 and decreases especially the take off weight.
Question 125-37 : An increase in pressure altitude ?
Generally reduces performance in class 1 and especially the take off weight.
Question 125-38 : Assuming an engine failure has occurred during take off, a performance class 1 helicopter which plans a turn in the climb of more than 15° must be capable of clearing vertical obstructions by an extra ?
There is no additional margin to apply.
Cat.pol.h.205 take off......b the take off mass shall be such that.1 it is possible to reject the take off and land on the fato in case of the critical engine failure being recognised at or before the take off decision point tdp.2 the rejected take off distance required rtodrh does not exceed the rejected take off distance available rtodah and.3 the todrh does not exceed the take off distance available todah..4 notwithstanding b 3 , the todrh may exceed the todah if the helicopter, with the critical engine failure recognised at tdp can, when continuing the take off, clear all obstacles to the end of the todrh by a vertical margin of not less than 10,7 m 35 ft...cat.pol.h.210 take off flight path......a from the end of the todrh with the critical engine failure recognised at the tdp.1 the take off mass shall be such that the take off flight path provides a vertical clearance, above all obstacles located in the climb path, of not less than 10,7 m 35 ft for operations under vfr and 10,7 m 35 ft + 0,01 x distance dr for operations under ifr. only obstacles as specified in cat.pol.h.110 have to be considered...2 where a change of direction of more than 15° is made, adequate allowance shall be made for the effect of bank angle on the ability to comply with the obstacle clearance requirements. this turn is not to be initiated before reaching a height of 61 m 200 ft above the take off surface unless it is part of an approved procedure in the afm.
Question 125-39 : Assuming one engine inoperative in a helicopter with performance class 1, during an approach to land, the following minimum rate of climb must be achievable ?
100 ft/min at 61 m.
Cs 29.67 climb one engine inoperative. oei. a for category a rotorcraft, in the critical take off configuration existing along the take off path, the following apply.. 1 the steady rate of climb without ground effect, 61 m 200 ft above the take off surface, must be at least 30 m 100 ft per minute, for each weight, altitude, and temperature for which take off data are to be scheduled with
Question 125-40 : Assuming that an engine fails at some point during take off, a helicopter of performance class 1 must be able to clear all obstacles, vertically, by ?
10.7 m + 0.01 dr in ifr.
Cat.pol.h.210 take off flight path..a from the end of the todrh with the critical engine failure recognised at the tdp.1 the take off mass shall be such that the take off flight path provides a vertical clearance, above all obstacles located in the climb path, of not less than 10,7 m 35 ft for operations under vfr and 10,7 m 35 ft + 0,01 x distance dr for operations under ifr. only obstacles as specified in cat.pol.h.110 have to be considered...2 where a change of direction of more than 15° is made, adequate allowance shall be made for the effect of bank angle on the ability to comply with the obstacle clearance requirements. this turn is not to be initiated before reaching a height of 61 m 200 ft above the take off surface unless it is part of an approved procedure in the afm.
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